TITLE: What does a linear transformation of $\mathbf{C}^2$ look like?
QUESTION [6 upvotes]: This is a math question, but is based on a physics topic I'm trying to develop an intuition for. Let me know if this question would be better placed in physics. 
What I know:
The "characteristic matrix" in thin film optics relates the E and H field amplitudes before and after passing through a thin film of material. For a single layer on a substrate, the equation is:  
$$
    \begin{pmatrix} 
    E_b \\ H_b 
    \end{pmatrix} =
    \begin{pmatrix}
    \cos(\delta) & i(1/y_1)\sin(\delta) \\
    iy_1 \sin(\delta) &  \cos(\delta) \\
    \end{pmatrix} 
     \begin{pmatrix} 
     E_a \\ H_a 
     \end{pmatrix}
$$
where $E_a,H_a$ are the field amplitudes at the film/substrate interface, $E_b,H_b$ are the field amplitudes at the air/film interface, $\delta$ = the phase thickness of the material, $y_1$ = characteristic admittance of the thin film. It can also be written in terms of the substrate admittance by normalizing to $E_a$:
$$
    \begin{pmatrix} 
    B \\ C 
    \end{pmatrix} =
    \begin{pmatrix}
    \cos(\delta) & i(1/y_1)\sin(\delta) \\
    iy_1 \sin(\delta) &  \cos(\delta) \\
    \end{pmatrix} 
     \begin{pmatrix} 
     1 \\ y_{sub} 
     \end{pmatrix}
$$
I was playing around with the matrix, when I realized: if you view $M$ as a linear transformation of $\mathbf{C}$, it corresponds to a pure rotation and scaling of the basis vectors (1,0) and (0,$ i$). In other words, for vector $\mathbf{v}$ in the complex plane, $M\mathbf{v}$ would rotate $\mathbf{v}$ by $\delta$ and then it's components would both be scaled by $y_1$ and $(1/y_1)$. See below for how the basis vectors rotate while staying normal to one another for $\delta = \frac{\pi}{2}$: 
$$
    \begin{pmatrix} 
    0 \\ iy_1 
    \end{pmatrix} =
    \begin{pmatrix}
    0 & i(1/y_1) \\
    iy_1 &  0 \\
    \end{pmatrix} 
     \begin{pmatrix} 
     1 \\ 0 
     \end{pmatrix}
$$
$$
    \begin{pmatrix} 
    -(1/y_1) \\ 0
    \end{pmatrix} =
    \begin{pmatrix}
    0 & i(1/y_1) \\
    iy_1 &  0 \\
    \end{pmatrix} 
     \begin{pmatrix} 
     0 \\ i 
     \end{pmatrix}
$$
I thought this was a really interesting geometric interpretation of what this matrix was doing, until I realized that the vectors $(E_a, H_a)$ or $(1, Y_{sub})$ have complex components themselves and therefore don't exist in $\mathbf{C}$, but actually in $\mathbf{C}^2$ or $\mathbf{R}^4$. 
What I don't know: So my questions now are, what does a linear transformation look like in $\mathbf{C}^2$? Does my pure rotation + scaling idea still make sense for this matrix? I'm not sure I even understand what $\mathbf{C}^2$ looks like much less how it would be transformed by a complex matrix...

REPLY [2 votes]: $\mathbf{C}^2$ is a bit of a mirage.   You may change your basis to
$$
    \begin{pmatrix} 
    E_b \\ H_b 
    \end{pmatrix} =
    \begin{pmatrix}
    \cos(\delta) & i(1/y_1)\sin(\delta) \\
    iy_1 \sin(\delta) &  \cos(\delta) \\
    \end{pmatrix} 
     \begin{pmatrix} 
     E_a \\ H_a 
     \end{pmatrix} \qquad \mapsto \qquad  \\  \begin{pmatrix} 
    E_b \\ iH_b 
    \end{pmatrix} =
    \begin{pmatrix}
    \cos(\delta) & (1/y_1)\sin(\delta) \\
    -y_1 \sin(\delta) &  \cos(\delta) \\
    \end{pmatrix} 
     \begin{pmatrix} 
     E_a \\ iH_a 
     \end{pmatrix},
$$
so the transition matrix is , in fact, a real rotation and scaling, unimodular,
 $$
 \begin{pmatrix}
 1/\sqrt{  y_1 } &  0 \\
     0 &  \sqrt{y_1}   \\
    \end{pmatrix} \begin{pmatrix}
    \cos(\delta) &  \sin(\delta) \\
    -  \sin(\delta) &  \cos(\delta) \\
    \end{pmatrix}  \begin{pmatrix}
    \sqrt{y_1 } &  0 \\
     0 & 1/ \sqrt{y_1 }  \\
    \end{pmatrix} .
$$